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Math Help - Solving Derivatives

  1. #1
    Senior Member vaironxxrd's Avatar
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    Solving Derivatives

    Hello Everyone,

    In preparation for an upcoming exam I've selected a derivative problem and attempted to solve it...

    I have to use f^' (x) = \frac{f(x+h)-f(x)}{h} to solve these problems.
    My attempts:

    a) f(x) = 3x^3 Using the book theorems I could quickly find the answer was 9x^2 Now I have to prove this...

    f^' (x) = \frac{3(x+h)^3-3x^3}{h} =

    \frac{3h^3+9h^2x+9hx^2+3x^3}{h} = (cancel h)

    9x^2+3x^3

    This solution of course doesn't match 9x^2
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  2. #2
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    Re: Solving Derivatives

    Quote Originally Posted by vaironxxrd View Post
    Hello Everyone,

    In preparation for an upcoming exam I've selected a derivative problem and attempted to solve it...

    I have to use f^' (x) = \frac{f(x+h)-f(x)}{h} to solve these problems.
    My attempts:

    a) f(x) = 3x^3 Using the book theorems I could quickly find the answer was 9x^2 Now I have to prove this...

    f^' (x) = \frac{3(x+h)^3-3x^3}{h} =

    \frac{3h^3+9h^2x+9hx^2+3x^3}{h} = (cancel h)

    9x^2+3x^3

    This solution of course doesn't match 9x^2
    It should be:
    \frac{3h^3+9h^2x+9hx^2}{h} = (cancel h)

    9x^2
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  3. #3
    Senior Member vaironxxrd's Avatar
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    Re: Solving Derivatives

    Quote Originally Posted by Plato View Post
    It should be:
    \frac{3h^3+9h^2x+9hx^2}{h} = (cancel h)

    9x^2
    I see I totally forgot the -3x^3.
    I see these (x+h)^n term are getting a little bit longer on some problems.
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